Question

The photoelectric work function for a metal surface is $4.125eV$ . The cutoff wavelength for this surface is.
A. $4125\stackrel{0}{A}$
B. $3000\stackrel{0}{A}$
C. $6000\stackrel{0}{A}$
D. $2062\stackrel{0}{A}$

Answer 1

Hint: In order to solve this question we need to understand the photoelectric effect. So in this effect, when light of some wavelength falls on a metal surface then it would knock out the electron from the metal surface, but for the electron to be removed, it has to supply energy so that the electron can overcome the potential barrier of metal, known as work function. Work function is mathematically expressed as a product of planck's constant and the threshold frequency or cut off frequency.

Complete step by step answer:
Let the threshold frequency be $v_0$ and planck's constant is given as $h=6.626\times {10}^{-34}Js$.According to problem, work function of metal is, $W=4.125eV$.
Since $1eV=1.6\times {10}^{-19}J$
So work function in joule is,
$W=4.125\times 1.6\times {10}^{-19}J$
$\Rightarrow W=6.6\times {10}^{-19}J$
So from definition of work function we get, $W=h{v}_0$
Frequency is given by, $v_0=Wh$
Putting values we get,
$v_0=6.6\times {10}^{-19}J6.626\times {10}^{-34}Js$
$\Rightarrow v_0=0.996\times {10}^{15}Hz$

Let the wavelength be, $\lambda$ and light speed in vacuum be $c=3\times {10}^{8}m{\sec }^{-1}$.
So, from definition we get, $\lambda v_0=c$
Wavelength is given by, $\lambda =c{v}_0$
Putting values we get,
$\lambda =3\times {10}^{8}m{\sec }^{-1}0.996\times {10}^{15}{\sec }^{-1}$
$\Rightarrow \lambda =3.01204\times {10}^{-7}m$
$\therefore \lambda =3000\stackrel{0}{A}$

So the correct option is B.

Note: It should be remembered that in photoelectric effect light is considered to be made of particles or energy balls known as photons. Particle nature of light was first established in blackbody radiation explanation and later it has been found that photoelectric effect, Compton scattering, pair production are all due to particle nature of light.